Question

1)2m²-2n²-4mn= 2)8p²-16+8= 3)12m⁵n+24m⁴n+12m³n= 4)(x²+2x)²= 5)81-(y²+6y)²=

Answer 1

Ответ:

(2m-2n)²

смотри на фото)дуиаю так

Answer 2

Ответ:

.................

Объяснение:

Answer 3

Let's tackle these step by step:

1. $2m^2 - 2n^2 - 4mn$

   • This is a quadratic expression. We can factor it by grouping: $2m^2 - 2n^2 - 4mn = 2m^2 - 2mn - 2mn - 2n^2$ $= 2m(m - n) - 2n(m - n)$ $= 2(m - n)(m + n)$

2. $8p^2 - 16 + 8$

   • First, simplify the constants: $-16 + 8 = -8$ $8p^2 - 8$
   • Factor out the common factor, which is 8: $8(p^2 - 1)$
   • Now, factor the difference of squares: $8(p - 1)(p + 1)$

3. $12m^5n + 24m^4n + 12m^3n$

   • Factor out the common factor, which is $12m^3n$: $12m^3n(m^2 + 2m + 1)$
   • Notice that $m^2 + 2m + 1$ is a perfect square trinomial: $(m + 1)^2$ $12m^3n(m + 1)^2$

4. $(x^2 + 2x)^2$

   • This is a square of a binomial. Square each term: $(x^2 + 2x)(x^2 + 2x)$ $= x^4 + 4x^3 + 4x^2$

5. $81 - (y^2 + 6y)^2$

   • This is a difference of squares: $a^2 - b^2 = (a + b)(a - b)$ $81 - (y^2 + 6y)^2 = (9 + y^2 + 6y)(9 - y^2 - 6y)$

There you go! Let me know if you have any questions about these solutions.